lim(x→π/4) (tanx) ^ tan2x (1+o)^∞ 类型,幂指函数,可先求其对数的极限.
令f(x) = (tanx) ^ tan2x,lnf(x) = tan2x ln(tanx) = ln(tanx) / (cot2x)
lim(x→π/4) ln(tanx) / (cot2x) 洛必达法则
= lim(x→π/4) (sec²x / tanx) / (-2 csc²2x)
= (1/2) /(-2) = -1/4
于是 lim(x→π/4) (tanx) ^ tan2x = e^(-1/4)